For all $x$ and $y$ in $[0,1]^2$
$$f(x,y)=
\left\{
\begin{aligned}
\int_x^y f_y(x,z)\,dz&\text{ if }x\le y, \\
-\int_y^x f_y(x,z)\,dz&\text{ if }x\ge y, 
\end{aligned}
\right.
$$
where $f_y(x,z):=\frac{\partial f(x,y)}{\partial y}|_{y=z}$, so that  
$$|f(x,y)|\le\int_0^1|f_y(x,z)|\,dz\le\sqrt{\int_0^1|f_y(x,z)|^2\,dz}. 
$$
Hence, 
$$1=\iint_{[0,1]^2}dx\,dy\,|f(x,y)|^2\le\iiint_{[0,1]^3}dx\,dy\,dz\,|f_y(x,z)|^2
=\iint_{[0,1]^2}dx\,dz\,|f_y(x,z)|^2
\le\|f-1\|_{H^1}^2. 
$$
So, any $c\in(0,1)$ will do.